I'm trying to calculate the extra drag created while the retractable landing lights are extended.
Ultimately I'm wanting to calculate the extra work and hence energy wasted by having these extended below 10,000 ft on descent.
I have found a website that can calculate force on various shapes given certain variables:
Drag Coefficient - DiracDelta Science & Engineering Encyclopedia
I'm assuming that the lights are hemisphere in shape and have a diameter of 0.27 m.
To keep things simple (and a little rough) I entered air density as ISA SL (1.225 kg/m^3)
Cd = 1.17
At 250 kt (125 m/s) it calculated the drag as 2564 Newtons.
2564/9.8= 261 kg.
Is that correct? It seems like a lot. I did a confidence check by entering the formula with 53 kt (100 km/h) and it came up with 119 N (12 kg) which seems reasonable if you imagine having such a shape out the car window and feeling the drag with your arm extended.
Anyway I was then wanting to compare the extra drag with total drag to convert it to a percentage of total drag.
If the aircraft has a L/D ratio of approx. 20:1 and the aircraft has a gross weight of say 65 tons (Max Landing Weight) the total drag would be approx. 3250 kg. So as a percentage, a single landing light extended at 250 kt would increase the drag by 8%. (261/3250). That doesn't sound right to me. I think it's out by a factor of ten but I can't see where the calculations go wrong.
I know the MEL increases the fuel burn by 1% per extended light so 0.8% sounds more like it.
And assuming a straight line descent from 10,000 ft to touchdown the distance travelled would be approximately 30 nm or 55.5 km (55,500 m)
Total energy in Joules is Newtons multiplied by distance (m). So according to the numbers above the amount of energy wasted is:
2564 N x 55,500 m = 142,302,000 Joules (142 MJ)
Can someone please confirm these numbers? Thank you.